| ⇦ | 
| ⇨ |
Electrons of mass m with de-Broglie wavelength λ fall on the target in an X-ray tube. The cut-off wavelength (λ₀) of the emitted X-ray is
Options
(a) λ₀= 2mcλ²/h
(b) λ₀= 2h/mc
(c) λ₀= 2m²c²λ³/h²
(d) λ₀= λ
Correct Answer:
λ₀= 2mcλ²/h
Explanation:
λ = h / p ⇒ p = h / λ
KE of electrons = E = p² / 2m = h² / 2mλ²
Also in X-ray, λ₀ = hc / E = 2mcλ² / h
Related Questions: - If a charge in current of 0.01 A in one coil produces a change in magnetic flux
- Two wires of the same dimensions but resistivities p1 and p2 are connected in series
- When a current of (2.5±0.5) A flows through a wire, it develops a potential difference
- A bottle full of water in a space ship at 30⁰C is carried on the moon.
- A circuit area 0.01 m² is kept inside a magnetic field which is normal to its plane.
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If a charge in current of 0.01 A in one coil produces a change in magnetic flux
- Two wires of the same dimensions but resistivities p1 and p2 are connected in series
- When a current of (2.5±0.5) A flows through a wire, it develops a potential difference
- A bottle full of water in a space ship at 30⁰C is carried on the moon.
- A circuit area 0.01 m² is kept inside a magnetic field which is normal to its plane.
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply